Finding the sum of the first $n$ terms of an AP

The sum of the (9)th to (17)th terms is (S_{17}-S_8=792). For a group of consecutive terms, take the difference of partial sums.

The governing property of partial sums is that Sₙ is the sum of the first n terms. Thus S₁₄ contains the first 14 terms, while S₆ contains the first 6 terms. Subtracting S₆ from S₁₄ removes the first six terms and leaves exactly the 7th through 14th terms: S₁₄ − S₆ = 665 − 165 = 500. Therefore option A is correct. No common difference or first term is needed because the required block is obtained directly from two partial sums. The other options arise from simple arithmetic errors in subtracting 165 from 665.

The sum of the (6)th to (12)th terms is (S_{12}-S_5=378). The sum of middle terms is found from the difference of partial sums.

The sum of the (10)th to (20)th terms is (1640-342=1298). For consecutive terms, use (S_m-S_k).

Putting (n=15) in (S_n=\frac{n}{2}[12+8(n-1)]) gives (930). Check the options in the formula.

Putting (n=16) in (S_n=\frac{n}{2}[20+9(n-1)]) gives (1240). The option method is fast when finding (n).

Putting (n=15) in (S_n=\frac{n}{2}[10+11(n-1)]) gives (1230). First identify (a) and (d) correctly.

Putting (n=16) in (S_n=\frac{n}{2}[8+12(n-1)]) gives (1504). Checking by substituting options is safe.

The numbers are (126,135,\ldots,297), and there are (20) terms, so the sum is (4230). In boundary questions, choose the first and last terms carefully.

The numbers are (112,126,\ldots,336), and there are (17) terms, so the sum is (3808). Find the first and last suitable multiples.

The numbers are (60,75,\ldots,240), and there are (13) terms, so the sum is (1950). Make the correct sequence by checking the limits.

The numbers are (208,224,\ldots,496), and there are (19) terms, so the sum is (6688). Do not forget to find the number of terms.

The sum of the first (n) odd numbers is (n^2), so (n^2=625) and (n=25). Recognizing perfect squares is useful.

The sum of the first (n) even numbers is (n(n+1)), so (27\times28=756). Do not confuse (n) with the last even number.

The sum of the first n odd natural numbers is n². Therefore, the sum of the first 26 odd numbers is 26² and that of the first 14 odd numbers is 14². Remaining sum = 26² − 14² = 676 − 196 = 480. A value such as 468 may result from using an incorrect average for the 12 remaining terms. Exam tip: Recall directly that the sum of the first n consecutive odd numbers is n².

The sum of the first 28 even natural numbers is \(28(28+1)=812\). The sum of the first 11 even natural numbers is \(11(11+1)=132\). Therefore, the sum of the remaining numbers is \(812-132=680\). A value such as 660 can result from an error in counting terms or subtraction. Exam tip: the sum of the first \(n\) even natural numbers is \(n(n+1)\).

Sum equals average (\times) number of terms, so (64\times15=960). If the average is given, the long formula is not necessary.

The formula for the average is \(\text{Average}=\frac{\text{sum of all terms}}{\text{number of terms}}\). Therefore, \(\frac{1170}{18}=65\). Hence, 65 is the correct answer. Option 64 is incorrect because \(18\times64=1152\), not the given sum of 1170. Exam tip: For an average question, divide the total sum by the number of terms.

The sum of the first \(n\) terms of an AP is \(S_n=\frac{n}{2}(a+l)\). Thus, \(912=\frac{12}{2}(18+l)=6(18+l)\). Hence, \(18+l=152\), so \(l=134\). If \(l=136\), the sum would be \(6(18+136)=924\), not the given sum. Exam tip: while finding the last term, first isolate \(a+l\) in the sum formula.

For an AP, the sum of the first \(n\) terms is \(S_n=\frac{n}{2}(a+l)\). Thus, \(975=\frac{15}{2}(a+97)\). This gives \(a+97=130\), so \(a=33\). If 35 were used as the first term, the sum would not be 975. Exam tip: When the last term is given, directly use \(S_n=\frac{n}{2}(a+l)\).